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Taya2010 [7]
3 years ago
13

Suppose 56%of politicians are lawyers.If a random sample of size 564 is selected, what is the probability that the proportion of

politicians who are lawyers will differ from the total politicians proportion by less than 4%
Mathematics
1 answer:
IceJOKER [234]3 years ago
8 0

Answer: 0.9444

Step-by-step explanation:

Given: The proportion of politicians are lawyers : <em>p </em>=0.56

Sample size : n = 564

Let q be th sample proportion.

The probability that the proportion of politicians who are lawyers will differ from the total politicians proportion by greater than 4% will be :-

P(|q-p|

Hence, the required probability = 0.9444

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Roses: 16
Total Cost: $20.64

Divide:
$20.64 ÷ 16
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Each Rose cost $1.29

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3 years ago
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Vesna [10]

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3 years ago
Helpppppp and explain tooo thank you :)
Wittaler [7]

Answer:

{5, 6, 7}

Step-by-step explanation:

When we have a given relation, the domain is the set of inputs, and the range as the set of the outputs.

so for a function f(x), and a domain {a. b. c}

The range is:

{f(a), f(b), f(c)}

In this case, we have:

f(x) = x + 6

and the domain is {-1, 0, 1}

Then the range is:

{ f(-1), f(0), f(1) }

{-1 + 6, 0 + 6, 1 + 6}

{5, 6, 7}

The correct option is the third one.

4 0
3 years ago
Can you help me with number 6? <br> Confused abit <br> Please
Sunny_sXe [5.5K]

You can see the three diagram attached. Each link is labeled with the probability: you have probability 1/6 that a six is rolled, and 5/6 that it is not rolled.


To answer the questions, find the path that brings you to the desired outcome, and multiply all the labels you meet.


First question:

To get three sixes, you have to choose the left path at each roll. The probability is always 1/6, so the answer is


\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} = \frac{1}{6^3}


Second question:

To get no sixes, you have to choose the right path at each roll. The probability is always 5/6, so the answer is


\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^3}{6^3}


Third question:

To get exactly one six, it can either be the first, second or third roll.


In all cases, you have to choose the left path once and the right path twice: left-right-right mean that you get the six in the first roll, right-left-right means that you get the six in the second roll, right-right-left means that you get the six in the third roll.


In every case, the left turn has probability 1/6, and the right turn has probability 5/6. The probability of each combination is thus


\frac{1}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{5^2}{6^3}


And since there are three of these combinations, The answer is


3\frac{5^2}{6^3}


Fourth question:

Since the question suggests to use what we already achieved, let's do it: having at least one six is the complementary event of having no sixes at all. If an event has probability p, its complementary has probability 1-p. So, since the probability of no sixes is known, the probability of at least one six is


1 - \frac{5^3}{6^3}

4 0
3 years ago
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